Question: $\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!}}~~$ is the Maclaurin series for which function? Choose 1 answer: Choose 1 answer: (Choice A) A $\sin x$ (Choice B) B $\sin \left( -x \right)$ (Choice C) C $\cos x$ (Choice D) D ${{e}^{x}}$ (Choice E) E ${{e}^{-x}}$
Solution: First note that all possible factorials appear in the denominators. This suggests that the given series most closely resembles the Maclaurin series for the function $~{{e}^{x}}$. ${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n}}}{n!}+...$ We need to get the signs of the terms to alternate in sign. Use $~-x~$ instead of $~x\,$. ${{e}^{-x}}=1-x+\frac{{{x}^{2}}}{2!}-...+\frac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!}+...~=~\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}{{x}^{n}}}{n!}}$